document.write( "Question 1081922: You roll a number cube twice. Find P(odd, then not 1). Write the probability as a fraction in simplest form.\r
\n" ); document.write( "\n" ); document.write( "So far I have P odd = 6/12 or 1/2 then I think (then not 1) is 6/12 - 2/12 = 4/12 or 1/3.\r
\n" ); document.write( "\n" ); document.write( "Am I correct?
\n" ); document.write( "

Algebra.Com's Answer #695964 by Fombitz(32388)\"\" \"About 
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\n" ); document.write( "First roll : odd.
\n" ); document.write( "Not sure why you have 6 out of 12.
\n" ); document.write( "There are only 6 possible outcomes.
\n" ); document.write( "Three of them are odd.
\n" ); document.write( "\"P%281%29=3%2F6=1%2F2\"
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\n" ); document.write( "Second roll : not a 1.
\n" ); document.write( "Again six possible outcomes, five are not 1.
\n" ); document.write( "\"P%282%29=5%2F6\"
\n" ); document.write( "So together,
\n" ); document.write( "\"P=P%281%29%2AP%282%29\"
\n" ); document.write( "\"P=%281%2F2%29%285%2F6%29\"
\n" ); document.write( "\"P=5%2F12\"
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